📚 Differentiation: From First Principles to Optimisation | 微分：从基本原理到最优化
Differentiation is one of the most powerful tools in A Level Mathematics. It measures how a quantity changes in response to another — the instantaneous rate of change. On the AQA specification, you are expected to differentiate confidently using a variety of rules and apply these skills to real-world optimisation problems. This article works through six core techniques, labelled (a) to (f), giving you a complete revision guide.
微分是A Level数学中最强大的工具之一。它衡量一个量相对于另一个量的变化——即瞬时变化率。在AQA考试大纲中，考生需要熟练运用多种法则进行微分，并将这些技能应用于现实世界的最优化问题。本文将通过六个核心技巧（编号 (a) 至 (f)）为你提供完整的复习指南。
1. (a) Understanding Gradients | 理解梯度
The gradient of a straight line is constant: it equals "rise over run". But for a curve, the gradient changes at every point. The derivative f' e tangent to the curve y = f(x) at any chosen x-value. Imagine zooming into a smooth curve — the curve looks more and more like a straight line. The gradient of this "infinitesimal straight line" is the derivative.
直线的梯度是恒定的：它等于"纵增量除以横增量"。但对于曲线，梯度在每一点都在变化。导数 f'(x) 给出曲线 y = f(x) 在任意选定 x 值处的切线斜率。想象放大一条光滑曲线——曲线看起来越来越像一条直线。这条"无穷小直线"的梯度就是导数。
The notation is important: dy/dx is Leibniz notation, while f' s Δx → 0. You should be comfortable switching between these two notations in exam answers.
记号很重要：dy/dx 是莱布尼茨记号，而 f'(x) 是拉格朗日记号。两者都会在AQA试卷中出现。当 x 增加一个微小量 Δx 时，y 的相应变化为 Δy。导数就是当 Δx → 0 时 Δy/Δx 的极限。你应该习惯在考试答案中自如切换这两种记号。
2. (b) Differentiation from First Principles | 从基本原理求导
AQA requires you to know the definition of the derivative from first principles. For a function f(x), the derivative at x is defined as the limit of the gradient of a chord joining two points that get closer and closer together:
AQA 要求你掌握导数从基本原理出发的定义。对于函数 f(x)，其在 x 处的导数定义为连接两个逐渐靠近的点的割线斜率的极限：
f'(x) = lim (h→0) [f(x+h) − f(x)] / h
dx (x²) = 2x. Notice that the h² term vanishes in the limit because h is approaching zero.
让我们将其应用于 f(x) = x²。首先，f(x+h) = (x+h)² = x² + 2xh + h²。减去 f(x) = x² 得到 2xh + h²。除以 h 得到 2x + h。当 h → 0 时，其趋于 2x。因此 d/dx (x²) = 2x。注意 h² 项在取极限时消失，因为 h 趋近于零。
The definition works for any polynomial, but the algebra can be lengthy. AQA often awards the full 4 or 5 marks for a first-principles question only if you write down the limit definition, substitute correctly, simplify fully, and safely take the limit. Missing the "lim" notation at any stage can cost a mark — always carry it through until the final step.
该定义适用于任何多项式，但代数运算可能较长。AQA 通常只有在你在第一原理题中写出极限定义、正确代入、完整化简并安全地取极限后，才会给满 4 或 5 分。在任何一步漏掉"lim"记号都可能扣一分——务必将其保留到最后一步。
3. (c) The Power Rule and Standard Results | 幂法则与标准结果
d/dx (x⁵) = 5x⁴, and d/dx (x⁻³) = −3x⁻⁴. The rule also works for fractional powers: d/dx (√x) = d/dx (x^½) = ½x−½ = 1/(2√x).
对于 f(x) = xⁿ（n 为任意实常数），幂法则指出 d/dx (xⁿ) = n xⁿ⁻¹。这是AQA微分题中使用频率最高的法则。例如，d/dx (x⁵) = 5x⁴，d/dx (x⁻³) = −3x⁻⁴。该法则同样适用于分数幂：d/dx (√x) = d/dx (x^½) = ½x−½ = 1/(2√x)。
x + 5) = 6x − 2. You must also know the derivatives of standard functions for AQA: d/dx (eˣ) = eˣ x) = cos x, and d/dx (cos x) = −sin x. The table below summarises these essential results.
常数可以提出来相乘：d/dx (7x³) = 21x²。和与差逐项微分：d/dx (3x² − 2x + 5) = 6x − 2。对于AQA，你还必须掌握标准函数的导数：d/dx (eˣ) = eˣ，d/dx (ln x) = 1/x，d/dx (sin x) = cos x，d/dx (cos x) = −sin x。下表总结了这些基本结论。
| f(x) | f'(x) |
| xⁿ | n xⁿ⁻¹ |
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | −sin x |
4. (d) The Product Rule | 乘积法则
oduct rule states: if y = uv, then dy/dx = u(dv/dx) + v(du/dx). In function notation: (fg)' = f'g + fg'. AQA examiners expect this layout: "u = ..., v = ..., du/dx = ..., dv/dx = ..." — it prevents errors and earns method marks.
当两个函数相乘时，我们不能简单地对每个因子分别求导。乘积法则指出：若 y = uv，则 dy/dx = u(dv/dx) + v(du/dx)。用函数记号表示：(fg)' = f'g + fg'。AQA 考官期望这样的排版："u = ...，v = ...，du/dx = ...，dv/dx = ..."——它可以避免错误并获得方法分。
Worked example: differentiate y = x²eˣ. Let u = x² and v = eˣ. Then du/dx = 2x and dv/dx = eˣ. Therefore dy/dx = x²eˣ + eˣ(2x) = eˣ(x² + 2x) = eˣ x(x + 2). Always factorise your final answer where possible — ised form, and an unfactorised answer may be marked wrong if it cannot be easily compared.
例题：求 y = x²eˣ 的导数。令 u = x²，v = eˣ。则 du/dx = 2x，dv/dx = eˣ。因此 dy/dx = x²eˣ + eˣ(2x) = eˣ(x² + 2x) = eˣ x(x + 2)。务必尽可能将最终答案因式分解——AQA 评分标准通常给出分解形式，若未分解的答案难以比对，可能被判错。
5. (e) The Quotient Rule | 商法则
y = u/v, then dy/dx = [v(du/dx) − u(dv/dx)] / v². The order in the numerator is critical: "low d-high minus high d-low over low squared" is a handy rhyme. Do not be tempted to reverse the subtraction — that is the most common sign error in this rule.
要对两个函数的分式求导，使用商法则：若 y = u/v，则 dy/dx = [v(du/dx) − u(dv/dx)] / v²。分子中的顺序至关重要："低乘高导减高乘低导，除以低平方
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